According to Bohr's theory, the ratio of electrostatic force of attraction acting on electron `3^(rd)` orbit of `He^(+)` ion and `2^(nd)` orbit of `Li^(2+)` ion is `((3)/(2))^(x)`. Then, the value of `x` is `:`

Options

A. 7
B. -6
C. 6
D. -7

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the electrostatic force formula
According to Bohr's theory, the electrostatic force of attraction (F) acting on an electron in a given orbit can be expressed as:
\[ F = \frac{k \cdot Z^2}{n^4} \]
where:
- \( k \) is a constant,
- \( Z \) is the atomic number of the ion,
- \( n \) is the principal quantum number (the orbit number).

### Step 2: Calculate the electrostatic force for the He\(^+\) ion in the 3rd orbit
For the He\(^+\) ion:
- \( Z = 2 \) (Helium has an atomic number of 2)
- \( n = 3 \)

Using the formula:
\[ F_{He^+} = \frac{k \cdot 2^2}{3^4} = \frac{k \cdot 4}{81} \]

### Step 3: Calculate the electrostatic force for the Li\(^{2+}\) ion in the 2nd orbit
For the Li\(^{2+}\) ion:
- \( Z = 3 \) (Lithium has an atomic number of 3)
- \( n = 2 \)

Using the formula:
\[ F_{Li^{2+}} = \frac{k \cdot 3^2}{2^4} = \frac{k \cdot 9}{16} \]

### Step 4: Find the ratio of the electrostatic forces
Now, we need to find the ratio of the forces:
\[ \text{Ratio} = \frac{F_{He^+}}{F_{Li^{2+}}} = \frac{\frac{k \cdot 4}{81}}{\frac{k \cdot 9}{16}} \]

This simplifies to:
\[ \text{Ratio} = \frac{4}{81} \cdot \frac{16}{9} = \frac{64}{729} \]

### Step 5: Express the ratio in terms of \(\left(\frac{3}{2}\right)^x\)
We know from the problem statement that:
\[ \frac{64}{729} = \left(\frac{3}{2}\right)^x \]

### Step 6: Rewrite \(64\) and \(729\) in terms of powers
We can express \(64\) and \(729\) as:
- \( 64 = 2^6 \)
- \( 729 = 3^6 \)

Thus, we can rewrite the ratio as:
\[ \frac{2^6}{3^6} = \left(\frac{2}{3}\right)^6 \]

### Step 7: Equate and solve for \(x\)
Now we have:
\[ \left(\frac{2}{3}\right)^6 = \left(\frac{3}{2}\right)^x \]

Taking the reciprocal of \(\frac{2}{3}\):
\[ \left(\frac{3}{2}\right)^{-6} = \left(\frac{3}{2}\right)^x \]

This implies:
\[ x = -6 \]

### Final Answer
Thus, the value of \(x\) is:
\[ \boxed{-6} \]